INEQUALITY FOR MONOTONIC MAPPINGS AND ITS APPLICATION FOR SOME SPECIAL MEANS
S. S. Dragomir
School of Communications and Informatics, Victoria University of Technology, P.O.Box 14428, MCMC, Melbourne, Victoria 8001, Australia
M. L. Fang†
Department of Mathematics, Nanjing Normal University, Nanjing 210097, P. R. China
Abstract. We first improve two Ostrowski type inequalities for monotonic func-tions, then provide its application for special means.
Keywords– Ostrowski’s Inequality, Trapezoid Inequality, Special Means.
1. Introduction.
In [1], Dragomir established the following Ostrowski’s inequality for monotonic mappings.
Theorem 1. Let f : [a, b]→R be a monotonic nondecreasing mapping on [a, b]. Then for all x∈[a, b], we have the following inequality
f(x)−
1 b−a
Z b
a
f(t)dt
≤
1 b−a
(
[2x−(a+b)]f(x) +
Z b
a
sgn(t−x)f(t)dt
)
≤b 1
−a[(x−a)(f(x)−f(a)) + (b−x)(f(b)−f(x))]
≤
1 2 +
|x−((a+b)/2)| b−a
(f(b)−f(a)). (1.1)
And the constant 1/2 is the best possible one.
In [2], Dragomir, Peˇcari´c and Wang generalized Theorem 1 and proved
†Supported in part by National Natural Science Foundation of China
Theorem 2. Let f : [a, b] →R be a monotonic nondecreasing mapping on [a, b] and t1, t2, t3 ∈(a, b) be such that t1 ≤t2 ≤t3. Then
Z b
a
f(x)dx−[(t1−a)f(a) + (t3−t1)f(t2) + (b−t3)f(b)]
≤(b−t3)f(b) + (2t2−t1−t3)f(t2)−(t1−a)f(a) +
Z b
a
T(x)f(x)dx
≤(b−t3)(f(b)−f(t3)) + (t3−t2)(f(t3)−f(t2))
+(t2−t1)(f(t2)−f(t1)) + (t1−a)(f(t1)−f(a))
≤max{t1−a, t2−t1, t3−t2, b−t3}(f(b)−f(a)), (1.2)
where T(x) =sgn(t1−x), for x ∈[a, t2], and T(x) =sgn(t3−x), for x∈[t2, b].
In the present paper, we firstly improve the above results, and then provide its application for some special means.
2. Main Result.
We shall start with the following result.
Theorem 3. Let f : [a, b]→R be a monotonic nondecreasing mapping on [a, b] and let t1, t2, t3 ∈[a, b] be such that t1 ≤t2 ≤t3. Then
Z b
a
f(x)dx−[(t1−a)f(a) + (t3−t1)f(t2) + (b−t3)f(b)]
≤max{(b−t3)(f(b)−f(t3)) + (t2−t1)(f(t2)−f(t1)),
(t3−t2)(f(t3)−f(t2)) + (t1 −a)(f(t1)−f(a))} (2.1)
≤max{t1−a, t2−t1, t3−t2, b−t3}(f(b)−f(a)). (2.2)
Proof. Since f(x) is a monotonic nondecreasing mapping on [a, b], we have
Z b
a
f(x)dx−[(t1−a)f(a) + (t3−t1)f(t2) + (b−t3)f(b)]
=
Z t1
a
(f(x)−f(a))dx+
Z t3
t1
(f(x)−f(t2))dx+
Z b
t3
(f(x)−f(b))dx
=
Z t1
a
(f(x)−f(a))dx+
Z t3
t2
(f(x)−f(t2))dx
− "Z t2
t1
(f(t2)−f(x))dx+
Z b
t3
(f(b)−f(x))dx#
≤max{(b−t3)(f(b)−f(t3)) + (t2−t1)(f(t2)−f(t1)),
(t3−t2)(f(t3)−f(t2)) + (t1−a)(f(t1)−f(a))}
≤max{t1−a, t2−t1, t3−t2, b−t3}(f(b)−f(a)).
Thus (2.1) and (2.2) are proved.
Corollary 1. Letf be defined as in Theorem 3. Then
Z b
a
f(x)dx−[(x−a)f(a) + (b−x)f(b)]
≤max{(b−x)(f(b)−f(x)),(x−a)(f(x)−f(a))}
≤max{x−a, b−x}max{(f(x)−f(a)),(f(b)−f(x))}
≤
1
2(b−a) +
x−
a+b 2
(f(b)−f(a)).
For x= (a+b)/2, we get trapezoid inequality.
Corollary 2. Let f be defined as in Theorem 3. Then
Z b
a
f(x)dx− f(a) +f(b)
2 (b−a)
≤b−2a max
f
a+b 2
−f(a)
,
f(b)−f
a+b 2
(2.3)
≤12(b−a)(f(b)−f(a)).
For t1 =a, t2 =x, t3 =b, we get Theorem 1.
3. Application for Special Means.
In this section, we shall give application of Corollary 2. Let us recall the fol-lowing means.
1. The arithmetic mean:
A =A(a, b) := a+b
2 , a, b ≥0. 2. The geometric mean:
G=G(a, b) :=√ab, a, b ≥0. 3. The harmonic mean:
H =H(a, b) := 2
1/a+ 1/b, a, b≥0. 4. The logarthmic mean:
L =L(a, b) := b−a
lnb−lna, a, b≥0, a6=b; If a=b, then L(a, b) =a. 5. The identric mean:
I =I(a, b) := 1 e
bb aa
1/(b−a)
6. The p-logarthmic mean:
Lp =Lp(a, b) :=
bp+1−ap+1 (p+ 1)(b−a)
1/p
, a 6=b; If a =b, then Lp(a, b) =a,
where p6=−1,0 anda, b > 0.
The following simple relationships are known in the literature
H ≤G≤L≤I ≤A.
We are going to use inequality (2.3) in the following equivalent version:
1 b−a
Z b
a
f(t)dt− f(a) +f(b) 2
≤1
2max
f
a+b 2
−f(a)
,
f(b)−f
a+b 2 )
(3.1)
≤1
2(f(b)−f(a)),
where f : [a, b]→Ris monotonic nondecreasing on [a, b].
5.1. Mapping f(x) =xp
Consider the mapping f : [a, b]⊂(0,∞)→R, f(x) =xp, p >0. Then
1 b−a
Z b
a
f(t)dt=Lpp(a, b),
f(a) +f(b)
2 =A(a
p, bp),
f(b)−f(a) =p(b−a)Lpp−−11. Then by (3.1), we get
Lpp(a, b)−A(ap, bp)≤1 2max
a+b 2
p
−ap, bp −
a+b 2
p
=1 2
bp −
a+b 2
p
= 1 2(b
p−ap)− 1
2
a+b 2
p −ap
≤21p(b−a)Lpp−−11− p(b−a)a
p−1
4 . (3.2)
Remark 1. The following result was proved in [2].
|Lpp(a, b)−A(ap, bp)| ≤ 1
2p(b−a)L
p−1
3.2. Mapping f(x) =−1/x
Consider the mapping f : [a, b]⊂(0,∞)→R, f(x) =−1/x. Then 1
b−a
Z b
a
f(t)dt=−L−1(a, b), f(a) +f(b)
2 =−
A(a, b) G2(a, b),
f(b)−f(a) = b−a G2(a, b).
Then by (3.1), we get
GA2((a, ba, b)) −L
−1(a, b)
≤12max
1 a −
2 a+b,
2 a+b−
1 b
=1 2
b−a a(a+b) =
1 2
b−a
ab −
1 2
b−a b(a+b)
≤1
2
b−a G2(a, b) −
1 2
b−a b(a+b). Thus we get
0≤AL−G2 ≤ 1 2
b
a+b(b−a)L. (3.3)
Remark 2. The following result was proved in [2]. 0≤AG−G2 ≤ 1
2(b−a)L.
3.3. Mapping f(x) = lnx
Consider the mapping f : [a, b]⊂(0,∞)→R, f(x) = lnx. Then 1
b−a
Z b
a
f(t)dt= lnI(a, b),
f(a) +f(b)
2 = lnG(a, b), f(b)−f(a) = b−a
L(a, b). Then by (3.1), we get
|lnI(a, b)−lnG(a, b)| ≤1 2max
lna+b
2 −lna,lnb−ln a+b
2
=1 2ln
a+b 2a =
1 2
b−a L(a, b) −
1 2 ln
2b a+b. Thus we get
1≤ I
G ≤
r
a+b 2b e
1
2Lb−a(a,b). (3.4)
Remark 3. The following result was proved in [2]. 1≤ I
G ≤e
References
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